Answer
(a) $25.71cm$ to the left of the concave lens
(b) $-0.38$
Work Step by Step
(a) The image formed by the first lens given as
$d_{i1}=\frac{(f_1)(d_{\circ 1})}{d_{\circ 1}-f_1}$
We plug in the known values to obtain:
$d_{i1}=\frac{(39.0cm)(400cm)}{400cm-39cm}$
$\implies d_{i1}=\frac{15600}{361}cm=43.2cm$
and the object distance of the second lens is $d_{\circ 2}=36.0cm-43.2cm=-7.2cm$
Now the image formed by the second lens is given as
$d_{i2}=\frac{(f_2)(d_{\circ 2})}{d_{\circ 2}-f_2}$
We plug in the known values to obtain:
$d_{i2}=\frac{(-10.0cm)(-7.2cm)}{-7.2cm+10.0cm}$
$\implies d_{i2}=25.71cm$ to the left of the concave lens.
(b) The magnification can be calculated as
$m=(\frac{-d_{i1}}{d_{\circ 1}})(\frac{-d_{i2}}{d_{\circ 2}})$
We plug in the known values to obtain:
$m=(\frac{-43.2cm}{400cm})(\frac{-25.71cm}{-7.2cm})$
$\implies m=-0.38$
